Answer:
electrical energy change into heat and light energy.
A low-luminosity star has a small and narrow <u>habitable zone</u>, whereas a high-luminosity star has a large and wide one.
<h3>What is luminosity of a star?</h3>
The radiant power emitted by a light-emitting item over time is measured as luminosity, which is an absolute measure of radiated electromagnetic power (light).
The total quantity of electromagnetic energy released per unit of time by a star, galaxy, or other celestial object is referred to as luminosity in astronomy.
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Answer:
a. 37.7 kgm/s b. 0.94 m/s c. -528.85 J
Explanation:
a. The initial momentum of block 1 of m₁ = 1.30 kg with speed v₁ = 29.0 m/s is p₁ = m₁v₁ = 1.30 kg × 29.0 m/s = 37.7 kgm/s
The initial momentum of block 2 of m₁ = 39.0 kg with speed v₂ = 0 m/s since it is initially at rest is p₁ = m₁v₁ = 39.0 kg × 0 m/s = 0 kgm/s
So, the magnitude of the total initial momentum of the two-block system = (37.7 + 0) kgm/s = 37.7 kgm/s
b. Since the blocks stick together after the collision, their final momentum is p₂ = (m₁ + m₂)v where v is the final speed of the two-block system.
p₂ = (1.3 + 39.0)v = 40.3v
From the principle of conservation of momentum,
p₁ = p₂
37.7 kgm/s = 40.3v
v = 37.7/40.3 = 0.94 m/s
So the final velocity of the two-block system is 0.94 m/s
c. The change in kinetic energy of the two-block system is ΔK = K₂ - K₁ where K₂ = final kinetic energy of the two-block system = 1/2(m₁ + m₂)v² and K₁ = final kinetic energy of the two-block system = 1/2m₁v₁²
So, ΔK = K₂ - K₁ = 1/2(m₁ + m₂)v² - 1/2m₁v₁² = 1/2(1.3 + 39.0) × 0.94² - 1/2 × 1.3 × 29.0² = 17.805 J - 546.65 J = -528.845 J ≅ -528.85 J
Answer:
Explanation:
An inelastic collision is one where 2 masses collide and stick together, moving as a single mass after the collision occurs. When we talk about this type of momentum conservation, the momentum is conserved always, but the kinetic momentum is not (the velocity changes when they collide). Because there is direction involved here, we use vector addition. The picture before the collision has the truck at a mass of 3520 kg moving north at a velocity of 18.5. The truck's momentum, then, is 3520(18.5) = 65100 kgm/s; coming at this truck is a car of mass 1480 kg traveling east at an unknown velocity. The car's momentum, then, is 1480v. The resulting vector (found when you pick up the car vector and stick the initial end of it to the terminal end of the truck's momentum vector) forms the hypotenuse of a right triangle where one leg is 65100 kgm/s, and the other leg is 1480v. Since we already know the final velocity of the 2 masses after the collision, we can use that to find the final momentum, which will serve as the resultant momentum vector in our equation (we'll get there in a sec). The final momentum of this collision is
p = mv and
p = (3520 + 1480)(13.6) so
p = 68000. Final momentum. The equation for this is a take-off of Pythagorean's Theorem and the one used to find the final magnitude of a resultant vector when you first began your vector math in physics. The equation is
which, in words, is
the final momentum after the collision is equal to the square root of the truck's momentum squared plus the car's momentum squared. Filling in:
and
and
and
and
so
v = 13.3 m/s at 72.6°
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